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$\frac{1}{1.3} + \frac{2}{1.3.5} +\frac {3}{1.3.5.7} + \frac{4}{1.3.5.7.9}........ n $ Terms.

I Know the answer to this problem but I couldn't find any proper way to actually solve this question. I thought the denominators were the product of n odd natural numbers. So I wrote the nth term as:

$\frac{2^{n+1}(n+1)!.n }{[2(n+1)]!}$

But I don't know if that's correct or even what to do next.?

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Aman_Bhaskar
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4 Answers4

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Hint: Note that:$$\frac{1}{3}=\frac{1}{2}\left(1-\frac{1}{3}\right);\frac{2}{15}=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{15}\right);\frac{3}{105}=\frac{1}{2}\left(\frac{1}{15}-\frac{1}{105}\right)...$$

Word Shallow
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the general term is given by $$a_n=\frac{n}{1*3*5....(2n+1)}$$

$$a_n=\frac{\frac{2n+1-1}{2}}{1*3*5....(2n+1)}$$ $$a_n=\frac{1}{2}*({\frac{2n+1}{1*3*5....(2n+1)}-\frac{1}{1*3*5...(2n+1)})}$$

$$a_n=\frac{1}{2}*({\frac{1}{1*3*5....(2n-1)}-\frac{1}{1*3*5...(2n+1)})}$$

for $n=1$ $$\frac{1}{2}*({\frac{1}{1}-\frac{1}{1*3})}$$ for $n=2$ $$\frac{1}{2}*({\frac{1}{1*3}-\frac{1}{1*3*5})}$$ for $n=3$ $$\frac{1}{2}*({\frac{1}{1*3*5}-\frac{1}{1*3*5*7})}$$

thus $$S_n=\frac{1}{2}*(1-\frac{1}{1*3*5*7*....(2n+1)}) $$

$$\lim_{n\to\infty} S_n=\frac{1}{2}$$

Deepesh Meena
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$$\sum_{k=1}^n\frac{k}{(2k+1)!!}=\frac{1}{2}\sum_{k=1}^n\left(\frac{1}{(2k-1)!!}-\frac{1}{(2k+1)!!}\right)=\frac{1}{2}\left(1-\frac{1}{(2n+1)!!}\right).$$

Michael Rozenberg
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The $n$th term has $n$ as its denominator, and the odd factors in $(2n+1)!$ as its numerator. Or, if we multiply both by the even factors, which are $\prod_{i=1}^n 2i=n!2^n$, the denominator becomes $(2n+1)!$. The $n$th term is therefore $\dfrac{n! n2^n}{(2n+1)!}$. (The formula you've obtained is also correct, by the way; you've just added a factor of $2(n+1)=2n+2$ into the two parts of the fraction.)

Alternatively, we could leave the numerator and denominator as they were originally, and write the $n$th term as $\dfrac{n}{(2n+1)!!}$, where the subfactorial $k!!$ is defined to only include the factors from $1$ to $k$ of the same parity as $k$. (In particular, the rescaling I suggested above used a factor of $(2n)!!$.)

As for the sum of the first $n$ terms, I think you can prove by induction it's $\frac{a_n}{2a_n+1}$ with $2a_n+1=(2n+1)!!$.

J.G.
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